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Update Ch 9 solutions
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‎ch09-lists-and-dictionaries/1-lists-multipurpose-containers.py

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‎ch09-lists-and-dictionaries/1-tuples-are-immutable-sequences.py

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# 9.1 - Tuples are Immutable Sequences
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# Solutions to review exercises
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# Exercise 1
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# Create a tuple "cardinal_numbers" with "first", "second" and "third"
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cardinal_numbers = ("first", "second", "third")
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# Exercise 2
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# Display the second object in the tuple
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print(cardinal_numbers[1])
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# Exercise 3
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# unpack the tuple into three string and display them
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position1, position2, position3 = cardinal_numbers
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print(position1)
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print(position2)
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print(position3)
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# Exercise 4
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# Create a tuple containing the letters of your name from a string
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my_name = tuple("David")
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# Exercide 5
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# Check whether or not x is in my_name
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print("x" in my_name)
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# Exercise 6
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# Create tuple containing all but the first letter in my_name
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print(my_name[1:])

‎ch09-lists-and-dictionaries/2-lists-are-mutable-sequences.py

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# 9.2 - Lists are Mutable Sequences
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# Solutions to review exercises
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# Exercise 1
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# Create a list named food with two elements "rice" and "beans".
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food = ["rice", "beans"]
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# Exercise 2
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# Append the string "broccolo" to the food list using .append()
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food.append("brocolli")
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# Exercise 3
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# Add the strings "bread" and "pizza" to food using .extend()
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food.extend(["bread", "pizza"])
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# Exercise 4
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# Print the first two items in food using slicing notation
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print(food[:2])
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# NOTE: The following is also acceptable
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print(food[0:2])
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# Exercise 5
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# Print the last item in food using index notation
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print(food[-1])
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# Exercise 6
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# Create a list called breakfast from 8000 the string "eggs, fruit, orange juice"
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breakfast = "eggs, fruit, orange juice".split(", ")
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# Exercise 7
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# Verify that breakfast has three items using len()
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print(len(breakfast) == 3)

‎ch09-lists-and-dictionaries/3-nesting-sorting-and-copying-lists-and-tuples.py

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# 9.3 - Nesting, Copying, and Sorting Lists and Tuples
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# Solutions to review exercises
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# Exercise 1
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# Create a tuple called data with two values, (1, 2) and (3, 4)
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data = ((1, 2), (3, 4))
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# Exercise 2
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# Loop over data a print the sum of each nested tuple
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for i in range(len(data)):
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print(f"Row {i+1} sum: {data[i][0] + data[i][1]}")
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# Exercise 3
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# Create the list [4, 3, 2, 1] and assign it to variable numbers
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numbers = [4, 3, 2, 1]
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# Exercise 4
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# Create a copy of the number list using [:]
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numbers_copy = numbers[:]
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# Exercise 5
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# Sort the numbers list in numerical order
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numbers.sort()
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print(numbers)

‎ch09-lists-and-dictionaries/2-challenge-list-of-lists.py renamed to ‎ch09-lists-and-dictionaries/4-challenge-list-of-lists.py

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# 9.2 - Challenge: List of Lists
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# 9.4 - Challenge: List of Lists
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# Solution to challenge
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‎ch09-lists-and-dictionaries/4-make-permanent-lists.py

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‎ch09-lists-and-dictionaries/3-challenge-wax-poetic.py renamed to ‎ch09-lists-and-dictionaries/5-challenge-wax-poetic.py

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# 9.3 - Challenge: Wax Poetic
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# 9.5 - Challenge: Wax Poetic
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# Solution to challenge
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‎ch09-lists-and-dictionaries/5-store-relationships-in-dictionaries.py renamed to ‎ch09-lists-and-dictionaries/6-store-relationships-in-dictionaries.py

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# 9.5 - Store Relationships in Dictionaries
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# 9.6 - Store Relationships in Dictionaries
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# Solutions to review exercises
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‎ch09-lists-and-dictionaries/6-challenge-capital-city-loop.py renamed to ‎ch09-lists-and-dictionaries/7-challenge-capital-city-loop.py

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# 9.6 - Challenge: Capital City Loop
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# 9.7 - Challenge: Capital City Loop
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# Solution to challenge
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‎ch09-lists-and-dictionaries/7a-challenge-cats-with-hats.py renamed to ‎ch09-lists-and-dictionaries/8a-challenge-cats-with-hats.py

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# 9.7 - Challenge: Cats With Hats
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# 9.8 - Challenge: Cats With Hats
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# Solution to challenge
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‎ch09-lists-and-dictionaries/7b-challenge-cats-with-hats.py renamed to ‎ch09-lists-and-dictionaries/8b-challenge-cats-with-hats.py

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# 9.7 - Challenge: Cats With Hats
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# 9.8 - Challenge: Cats With Hats
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# Alternative solution to challenge
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‎ch09-lists-and-dictionaries/7c-challenge-cats-with-hats.py renamed to ‎ch09-lists-and-dictionaries/8c-challenge-cats-with-hats.py

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# 9.7 - Challenge: Cats With Hats
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# 9.8 - Challenge: Cats With Hats
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# Alternative solution to challenge using dictionaries
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